Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 16 - Planar Kinematics of a Rigid Body - Section 16.4 - Absolute Motion Analysis - Problems - Page 343: 43



Work Step by Step

We can determine the required angular velocity as follows: $sin\phi=\frac{x}{0.3}$ $\implies x=0.3sin\phi$ and $cos\phi=\frac{y}{0.3}$ $y=0.3cos\phi$ Thus, $tan\theta=\frac{x}{0.6-y}$ $\implies tan\theta=\frac{0.3sin\phi}{0.6-0.3cos\phi}$ $\implies 0.3sin\phi=tan\theta(0.6-0.3cos\phi)$ $\implies 0.3sin\phi=0.6tan\theta-0.3cos\phi tan\theta$ Differentiating with respect to time, we obtain: $0.3cos\phi\frac{d\phi}{dt}=0.6sec^2\frac{d\theta}{dt}-0.3cos\phi sec^2\theta \frac{d\theta}{dt}+0.3sin\phi tan\theta \frac{d\phi}{dt}$ We plug in the known values to obtain: $0.3cos60\omega_{AB}=0.6sec^2 30\omega_{CD}-0.3cos60 sec^2 30\omega_{CD}+0.3sin60tan30\omega_{AB}$ $0.3cos60(4)=0.6sec^2 30\omega_{CD}-0.3cos60sec^230\omega_{CD}+0.3sin60tan30(4)$ This simplifies to: $\omega_{CD}=0rad/s$
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